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Math Help - Transformation of uniform random variables: Stuck on a question about density functio

  1. #1
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    Transformation of uniform random variables: Stuck on a question about density functio

    Given: X and Y, independent, continuous uniform random variables on the interval (0,1)

    Goal: Find the density function of  \frac{X}{X+Y}

    My approach:
    Use the transformation technique (as in point 5 of this wikipedia page Integration by substitution - Wikipedia, the free encyclopedia).

    Define $Z_{1} = \frac{X}{X+Y}$. Define $Z_{2} = X$.
    I find that the joint density of  $Z_{1}$ and $Z_{2}$ is

     $f_{Z_{1},Z_{2}}(z_{1},z_{2}) = \frac{Z_{2}}{{Z_{1}}^{2}}$

    And this is where i get stuck... I tried integrating Z_{2} out of it but nothing reasonable comes out (not a density function anyway). So two questions:

    1. Did I find the joint density correct?
    2. I tried integrating  Z_{2} over the interval  Z_{1} to 1, is this correct?
    3. What should be next step?

    Thanks for any help!
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: Transformation of uniform random variables: Stuck on a question about density fun

    Quote Originally Posted by celestial View Post
    Given: X and Y, independent, continuous uniform random variables on the interval (0,1)

    Goal: Find the density function of  \frac{X}{X+Y}

    My approach:
    Use the transformation technique (as in point 5 of this wikipedia page Integration by substitution - Wikipedia, the free encyclopedia).

    Define $Z_{1} = \frac{X}{X+Y}$. Define $Z_{2} = X$.
    I find that the joint density of  $Z_{1}$ and $Z_{2}$ is

     $f_{Z_{1},Z_{2}}(z_{1},z_{2}) = \frac{Z_{2}}{{Z_{1}}^{2}}$

    And this is where i get stuck... I tried integrating Z_{2} out of it but nothing reasonable comes out (not a density function anyway). So two questions:

    1. Did I find the joint density correct?
    2. I tried integrating  Z_{2} over the interval  Z_{1} to 1, is this correct?
    3. What should be next step?

    Thanks for any help!
    May be that it is useful to devide the procedure into two steps. The first step is to find the p.d.f. of the random variable Z=\frac{X}{Y} being X and Y independent random variables uniformely distributed in [0,1]. Proceeding as in...

    http://www.mathhelpforum.com/math-he...es-193906.html

    ... we find for 0<z<1...

    p_{z}(z)= \int_{0}^{1} y\ dy =\frac{1}{2} (1)

    ... for z>1...

    p_{z}(z)= \int_{0}^{\frac{1}{z}} y\ dy =\frac{1}{2\ z^{2}} (2)

    ... and for z<0...

    p_{z}(z)= 0 (3)

    The second step is to find the p.d.f. of the random variable U=\frac{1}{1+Z}... how to proceed?...



    Marry Christmas from Serbia

    \chi \sigma
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    Re: Transformation of uniform random variables: Stuck on a question about density fun

    Quote Originally Posted by chisigma View Post
    May be that it is useful to devide the procedure into two steps. The first step is to find the p.d.f. of the random variable Z=\frac{X}{Y} being X and Y independent random variables uniformely distributed in [0,1]. Proceeding as in...

    http://www.mathhelpforum.com/math-he...es-193906.html

    ... we find for 0<z<1...

    p_{z}(z)= \int_{0}^{1} y\ dy =\frac{1}{2} (1)

    ... for z>1...

    p_{z}(z)= \int_{0}^{\frac{1}{z}} y\ dy =\frac{1}{2\ z^{2}} (2)

    ... and for z<0...

    p_{z}(z)= 0 (3)

    The second step is to find the p.d.f. of the random variable U=\frac{1}{1+Z}... how to proceed?...



    Marry Christmas from Serbia

    \chi \sigma
    Nice trick with the U! Thanks for that. I knew the ratio distribution of two standard uniforms but couldn't get any use if it. So let me see if Mary and Jesus have answered my prayers and blessed me with knowledge:

    So using the ratio of two uniform variables, let's call g(z) = M = \frac{1}{1+Z}. Assume we can prove all the desirable properties like g being a bijective function, and its inverse having a continuous derivative (I haven't proven them yet), then I have g^{-1}(M) = \frac{1-M}{M}. Hence I have for f_{M}(m) =  \frac{ \partial g^{-1}(m)}{\partial m}f_{z}(g^{-1}(m))
    So then I'll have

    for  0 < \frac{1-m}{m} < 1
    • f_{M}(m) = - \frac{1}{2 \cdot m^{2}}

    for   \frac{1-m}{m} > 1

    • f_{M}(m) = - \frac{1}{2 \cdot m^{2} \cdot \left(\frac{1-m}{m}\right)}

    and 0 otherwise?


    Of course I need to simplify the conditions on m (the intervals) , but is that the right direction?
    Last edited by celestial; December 12th 2011 at 12:59 PM.
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  4. #4
    MHF Contributor chisigma's Avatar
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    Re: Transformation of uniform random variables: Stuck on a question about density fun

    A preminary investigation on the functional equation...

    U=\frac{1}{1+Z},\ Z \ge 0 (1)

    ... may be useful. From (1) it is clear that for u<0 is P\{U<u\}=0 and for u>0 is P\{U<u\}=1, so that we have to investigate what happens for 0<u<1. From (1) we derive...

    P\{U<u\} = P\{Z>\frac{1-U}{U}\}= \int_{\frac{1-u}{u}}^{\infty} p_{z}(z)\ dz (2)

    The function p_{z}(*) has been computed in a previous post...

    p_{z}(z)=\begin{cases}\frac{1}{2} &\text{if}\ 0<z<1\\ \frac{1}{2\ z^{2}} &\text{if}\ z>1\\ 0 &\text{elsewhere}\end{cases} (3)

    Now deriving (2) and taking into account (3) we obtain...

    p_{u}(u)=\begin{cases}\frac{1}{2\ (1-u)} &\text{if}\ 0<z<\frac{1}{2}\\ \frac{1}{2\ u^{2}} &\text{if}\ \frac{1}{2}<z<1\\ 0 &\text{elsewhere}\end{cases} (4)




    Marry Christmas from Serbia

    \chi \sigma
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